Question: If $x \circledcirc y = x(y-6)$ and $x \odot y = x-7y$, find $(0 \circledcirc -3) \odot -3$.
First, find $0 \circledcirc -3$ $ 0 \circledcirc -3 = 0$ $ \hphantom{0 \circledcirc -3} = 0$ Now, find $0 \odot -3$ $ 0 \odot -3 = 0-(7)(-3)$ $ \hphantom{0 \odot -3} = 21$.